LeetCode(Merge Two Sorted Lists ) 合并两个有序的链表

最新推荐文章于 2025-07-23 10:11:23 发布

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#面试 #leetcode #递归 #merge #链表

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本文深入探讨了使用递归和循环方法合并两个已排序的链表，并详细解释了算法实现过程和核心逻辑。通过实例演示，帮助读者理解如何高效地将两个有序链表整合为一个新的有序链表。

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题目要求：

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

这道题可以用递归和循环的方式完成：

代码：

  ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
    if(l1 == NULL)
      return l2;
    if(l2 == NULL)
      return l1;
    ListNode* merge_head = NULL;
    if(l1->val < l2->val)
    {
      merge_head = l1;
      l1 = l1->next;
    }
    else 
    {
      merge_head = l2;
      l2 = l2->next;
    }
    ListNode* cur = merge_head;
    while(l1 != NULL && l2 != NULL)
    {
      if(l1->val < l2->val)
      {
        cur->next = l1;
        l1 = l1->next;
      }
      else 
      {
        cur->next = l2;
        l2 = l2->next;
      }
      cur = cur->next;
    }
    while(l1 != NULL)
    {
      cur->next = l1;
      l1 = l1->next;
      cur = cur->next;
    }
    while(l2 != NULL)
    {
      cur ->next = l2;
      l2 = l2->next;
      cur = cur->next;
    }
    return merge_head;
  }
  
  ListNode* MergeRecursive(ListNode* l1, ListNode* l2)
  {
    if(l1 == NULL)
      return l2;
    if(l2 == NULL)
      return l1;
    ListNode* merge_head = NULL;
    if(l1->val < l2->val)
    {
      merge_head = l1;
      merge_head->next = MergeRecursive(l1->next, l2);
    }
    else 
    {
      merge_head = l2;
      merge_head->next = MergeRecursive(l1, l2->next);
    }
    return merge_head;
  }